In Internet communication field, such as VoIP or IP camera, in order to assure that the users are online continually so as to make both sides connect with each other, the users are requested to perform registration intermittently to confirm interconnection.
Referring to FIG. 1, if PC1 and PC2 are going to conduct interconnection with each other through VoIP or IP camera, both sides must perform registrations intermittently to the server 3 through NAT 4 or NAT 5 respectively. The server 3 will then respond to PC1 and PC2 respectively to express that interconnection is achieved (step {circle around (1)} registration packet to and fro). When PC 2 is going to connect with PC 1, an Invite packet is sent to the server 3 through NAT 5 (step {circle around (2)}), then the server 3 sends the Invite packet to PC 1 through NAT 4 (step {circle around (3)}). PC1 will send a responding packet to the server 3 through NAT 4 (step {circle around (4)}), then the server 3 sends the responding packet to PC 2 through NAT 5 (step {circle around (5)}).
Referring to FIG. 2, the server 3 is not only faced by PC 1 and PC 2, but also faced by thousands of user. In order to reduce the load of the server 3 for receiving registrations continuously, it is a natural development in Internet communication that the users are requested to increase the time interval of registration intermittently. But if NAT 4 and NAT 5 find that no packet is passed through for a long time (T2 time), then NAT 4 and NAT 5 will shut down automatically, while the server 3, PC 1 and PC 2 are all ignorant of the status. Therefore the Invite packets in step {circle around (2)} or step {circle around (3)} cannot be sent, and must wait till the next time PC1 or PC2 sends a registration packet to open NAT 4 or NAT 5.
Referring to FIG. 3, the interactions between PC 1 and the server 3 are used as an example for description, the interactions between PC 2 and the server 3 can be applied by analogy. When PC 1 sends a registration forward packet, T1 time information is carried for sending to the server 3 through NAT 4. The server 3 then responds with a registration backward packet, and knows that PC 1 will register again after T1 time.
PC 1 waits for T1 time, then sends the second registration forward packet, a T3 time information is carried for sending to the server 3 through NAT 4, T3>T2. The server 3 then responds with a registration backward packet, and knows that PC 1 will register again after T3 time.
The shutdown time for NAT 4 is T2, T1<T2<T3. If NAT 4 itself finds that no packet is passed through within T2 time, then NAT 4 will shutdown automatically, while the server 3, PC 1 are ignorant of the status. At this time, if PC 2 sends an Invite packet to the server 3 through NAT 5 after T2 time, the Invite packet will be blocked up by NAT 4 when the Invite packet is sent by the server 3 to PC 1 through NAT 4.
After T3 time, PC 1 sends the third registration forward packet so as to open NAT 4, a T4 time information is carried for sending to the server 3 through NAT 4, T4>T3. Therefore a vicious circle is formed so that PC 2 cannot send the Invite packet to PC 1 through NAT 4.